Problem: $\dfrac{dy}{dx}=\sqrt{xy}$ Which curve solves the differential equation and passes through the point $(0,9)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y=x^3+9$ (Choice B) B $y=\dfrac19x^3+81$ (Choice C) C $y=\dfrac19x^3+9$ (Choice D) D $y=\dfrac19x^3+2x^{3/2}+81$ (Choice E) E $y=x^3+81$ (Choice F) F $y=\dfrac19x^3+2x^{3/2}+9$
Explanation: The differential equation is separable. $\dfrac{dy}{dx}=\sqrt x\cdot\!\!\sqrt y$ What does it look like after we separate the variables? $y^{-1/2}\,dy=x^{1/2}\,dx$ Let's integrate both sides of the equation. $\int y^{-1/2}\,dy=\int x^{1/2}\,dx$ What do we get? $2y^{1/2}=\dfrac23x^{3/2} + C$ What value of $C$ makes the solution curve pass through the point $(0,9)$ ? Let's substitute $x=0$ and $y=9$ into the equation and solve for $C$. $\begin{aligned} 2\cdot9^{1/2} &= \dfrac23\cdot 0^{3/2} + C\\ \\ \\ 2\cdot3&=0+C\\ \\ C&=6 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} 2y^{1/2}&=\dfrac23x^{3/2} + 6\\ \\ \\ y^{1/2}&=\dfrac13x^{3/2} + 3\\ \\ \\ y&=\left(\dfrac13x^{3/2} + 3\right)^2\\ \\ \\ \\ y&=\dfrac19x^3+2x^{3/2}+9 \end{aligned}$